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Chapter 3 GRAVITY DAMS Basically, gravity dams are solid concrete structures that maintain their stability against design loads from the geometric shape and the weight and strength of the structure. Generally, they are constructed on a straight axis, but may be slightly curved or angled to accommodate the specific site conditions. Gravity dams typically consist of a non-overflow section(s) and an overflow section or spillway. They are constructed with masonry or concrete but of late conventional concrete or roller-compacted concrete are popular. The two general concrete construction methods for concrete gravity dams are conventional placed mass concrete and roller-compacted concrete (RCC). a. Conventional concrete dams. (1) Conventionally placed mass concrete dams are characterized by construction using materials and techniques employed in the proportioning, mixing, placing, curing, and temperature control of mass concrete. Construction incorporates methods that have been developed and perfected over many years of designing and building mass concrete dams. The cement hydration process of conventional concrete limits the size and rate of concrete placement and necessitates building in monoliths to meet crack control requirements. Generally using large-size coarse aggregates, mix proportions are selected to produce a low-slump concrete that gives economy, maintains good workability during placement, develops minimum temperature rise during hydration, and produces important properties such as strength, impermeability, and durability. Dam construction with conventional concrete readily facilitates installation of conduits, penstocks, galleries, etc., within the structure. (2) Construction procedures include batching and mixing, and transportation, placement, vibration, cooling, curing, and preparation of horizontal construction joints between lifts. The large volume of concrete in a gravity dam normally justifies an onsite batch plant, and requires an aggregate source of adequate quality and quantity, located at or within an economical distance of the project. Transportation from the batch plant to the dam is generally performed in buckets ranging in size from 4 to 12 cubic yards carried by truck, rail, cranes, cableways, or a combination of these methods. The maximum bucket size is usually restricted by the capability of effectively spreading and vibrating the concrete pile after it is dumped from the bucket. The concrete is placed in lifts of 5- to 10-foot depths. Each lift consists of successive layers not exceeding 18 to 20 inches. Vibration is generally performed by large one-man, air-driven, spud-type vibrators. Methods of cleaning horizontal construction joints to remove the weak laitance film on the surface during curing include green cutting, wet sand-blasting, and high-pressure air-water jet.

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3) The heat generated as cement hydrates requires careful temperature control during placement of mass concrete and for several days after placement. Uncontrolled heat generation could result in excessive tensile stresses due to extreme gradients within the mass concrete or due to temperature reductions as the concrete approaches its annual temperature cycle. Control measures involve precooling and postcooling techniques to limit the peak temperatures and control the temperature drop. Reduction in the cement content and cement replacement with pozzolans have reduced the temperature-rise potential. Crack control is achieved by constructing the conventional concrete gravity dam in a series of individually stable monoliths separated by transverse contraction joints. Usually, monoliths are approximately 50 feet wide. b. Roller-compacted concrete (RCC) gravity dams. The design of RCC gravity dams is similar to conventional concrete structures. The differences lie in the construction methods, concrete mix design, and details of the appurtenant structures. Construction of an RCC dam is a relatively new and economical concept. Economic advantages are achieved with rapid placement using construction techniques that are similar to those employed for embankment dams. RCC is a relatively dry, lean, zero slump concrete material containing coarse and fine aggregate that is consolidated by external vibration using vibratory rollers, dozer, and other heavy equipment. In the hardened condition, RCC has similar properties to conventional concrete. For effective consolidation, RCC must be dry enough to support the weight of the construction equipment, but have a consistency wet enough to permit adequate distribution of the past binder throughout the mass during the mixing and vibration process and, thus, achieve the necessary compaction of the RCC and prevention of undesirable segregation and voids. Site Selection a. General. During the feasibility studies, the preliminary site selection will be dependent on the project purposes. Purposes applicable to dam construction include navigation, flood damage reduction, hydroelectric power generation, fish and wildlife enhancement, water quality, water supply, and recreation. The feasibility study will establish the most suitable and economical location and type of structure. b. Selection factors. (1) A concrete dam requires a sound bedrock foundation. It is important that the bedrock have adequate shear strength and bearing capacity to meet the necessary stability requirements. The foundation permeability and the extent and cost of foundation grouting, drainage, or other seepage and uplift control measures should be investigated.

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(2) The topography is an important factor in the selection and location of a concrete dam and its appurtenant structures. Construction at a site with a narrow canyon profile on sound bedrock close to the surface is preferable, as this location would minimize the concrete material requirements and the associated costs. (3) The criteria set forth for the spillway, powerhouse, and the other project appurtenances will play an important role in site selection. The relationship and adaptability of these features to the project alignment will need evaluation along with associated costs. (4) Additional factors of lesser importance that need to be included for consideration are the relocation of existing facilities and utilities that lie within the reservoir and in the path of the dam. Included in these are railroads, powerlines, highways, towns, etc. Extensive and costly relocations should be avoided. (5) The method or scheme of diverting flows around or through the damsite during construction is an important consideration to the economy of the dam. A concrete gravity dam offers major advantages and potential cost savings by providing the option of diversion through alternate construction blocks, and lowers risk and delay if overtopping should occur. Forces on Gravity Dams 3-3. Loads a. General. In the design of concrete gravity dams, it is essential to determine the loads required in the stability and stress analysis. The following forces may affect the design: (1) Dead load. (2) Water Pressure (Headwater and tailwater pressures). (3) Uplift. (4) Temperature. (5) Earth and silt pressures. (6) Ice pressure. (7) Earthquake forces. (8) Wind pressure. (9) Subatmospheric pressure. (10) Wave pressure. (11) Reaction of foundation. b. Dead load. Dead load comprises the major resisting force. The dead loads considered should include the weight of concrete, superimposed backfill, and appurtenances such as gates and bridges. In the computation of the dead load, relatively small voids such as galleries are normally not deducted except in low dams, where such voids could create an appreciable effect upon the stability of the structure. The cross section of the dam is divided into several triangles and rectangles. Weight of each triangle and rectangle and their points of application at respective centre of gravity are computed. The resultant of all these downward forces is thus found by talking moments of the component forces which constitute the total weight of the dam

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acting at its centre of gravity. Unit weight of concrete and masonry is taken as 2400kg/m3 and 2300kg/m3. c. Water Pressure; (Headwater and tailwater). Water pressure is the major external force acting on the dam. As the water is stored in the reservoir, and stands against the body of the dam, it exerts horizontal pressure on the dam.

Fig. G.D. 1 Headwater pressure with vertical upstream face

Fig. G.D. 2 Water pressure for slanted upstream face and water at tailrace When the upstream face of the dam is vertical, the water pressure P = h2 and acting at h/3 from base. When the upstream face is slanted, the water pressure is resolved in two components: Horizontal oressure P = h2 and acting at h/3 from base.

P2

Tail water h

P1

Ww Ww

Head water

h

P=1/2h

h/3

Max. flow level

Dam

5

Vertical pressure Ww = weight of water on slanted side and acting at centre of gravity of volume of water. Uplift Pressure. It is the second major external force acting upwards on the dam. Uplift pressure resulting from headwater and tailwater exists through cross sections within the dam, at the interface between the dam and the foundation, and within the foundation below the base. This pressure is present within the cracks, pores, joints, and seams in the concrete and foundation material. Uplift pressure is an active force that must be included in the stability and stress analysis to ensure structural adequacy. These pressures vary with time and are related to boundary conditions and the permeability of the material. (1) Along the base. (a) General. The uplift pressure will be considered as acting over 100 percent of the base. A hydraulic gradient between the upper and lower pool is developed between the heel and toe of the dam. The pressure distribution along the base and in the foundation is dependent on the effectiveness of drains and grout curtain, where applicable, and geologic features such as rock permeability, seams, jointing, and faulting. The uplift pressure at any point under the structure will be tailwater pressure plus the pressure measured as an ordinate from tailwater to the hydraulic gradient between upper and lower pool.

Fig. G.D.3 Uplift distribution without foundation drainage (b) Without drains. Where there have not been any provisions provided for uplift reduction, the hydraulic gradient will be assumed to vary, as a straight line, from headwater at the heel to zero or tailwater at the toe. Determination of uplift, at any point on or below the foundation, is demonstrated in Figure G.D.3 above.

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(c) With drains. Uplift pressures at the base or below the foundation can be reduced by installing foundation drains. The effectiveness of the drainage system will depend on depth, size, and spacing of the drains; the character of the foundation; and the facility with which the drains can be maintained. This effectiveness will be assumed to vary from 25 to 50 percent, and the design memoranda should contain supporting data for the assumption used. Along the base, the uplift pressure will vary linearly from the undrained pressure head at the heel, to the reduced pressure head at the line of drains, to the undrained pressure head at the toe, as shown in Figure G.D.4

Figure G.D. 4 Uplift distribution with drainage gallery Where the line of drains intersects the foundation within a distance of 5 percent of the reservoir depth from the upstream face, the uplift may be assumed to vary as a single straight line, which would be the case if the drains were exactly at the heel. This condition is illustrated in Figure G.D.5 If the drainage gallery is above tailwater elevation, the pressure of the line of drains should be determined as though the tailwater level is equal to the gallery elevation.

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Figure G.D.5. Uplift distribution with foundation drains near upstream face (d) Grout curtain. For drainage to be controlled economically, retarding of flow to the drains from the upstream head is mandatory. This may be accomplished by a zone of grouting (curtain) or by the natural imperviousness of the foundation. A grouted zone (curtain) should be used wherever the foundation is amenable to grouting. Grout holes shall be oriented to intercept the maximum number of rock fractures to maximize its effectiveness. Under average conditions, the depth of the grout zone should be two-thirds to three-fourths of the headwater-tailwater differential and should be supplemented by foundation drain holes with a depth of at least two-thirds that of the grout zone (curtain).

Figure G.D.6. Uplift distribution cracked base with drainage, zero compression zone not extending beyond drains (3-4)

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Where the foundation is sufficiently impervious to retard the flow and where grouting would be impractical, an artificial cutoff is usually unnecessary. Drains, however, should be provided to relieve the uplift pressures that would build up over a period of time in a relatively impervious medium. In a relatively impervious foundation, drain spacing will be closer than in a relatively permeable foundation. (e) Zero compression zones. Uplift on any portion of any foundation plane not in compression shall be 100 percent of the hydrostatic head of the adjacent face, except where tension is the result of instantaneous loading resulting from earthquake forces. When the zero compression zone does not extend beyond the location of the drains, the uplift will be as shown in Figure G.D.6. For the condition where the zero compression zone extends beyond the drains, drain effectiveness shall not be considered. This uplift condition is shown in Figure G.D.7

Figure G.D.7. Uplift distribution cracked base with drainage, zero compression zone extending beyond drains(3-5) . When an existing dam is being investigated, the design office should submit a request to CECW-ED for a deviation if expensive remedial measures are required to satisfy this loading assumption. (2) Within dam. (a) Conventional concrete. Uplift within the body of a conventional concrete-gravity dam shall be assumed to vary linearly from 50 percent of maximum headwater at the upstream face to 50 percent of tailwater, or zero, as the case may be, at the downstream face. This simplification is based on the relative

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impermeability of intact concrete which precludes the buildup of internal pore pressures. Cracking at the upstream face of an existing dam or weak horizontal construction joints in the body of the dam may affect this assumption. (b) RCC concrete. The determination of the percent uplift will depend on the mix permeability, lift joint treatment, the placements, techniques specified for minimizing segregation within the mixture, compaction methods, and the treatment for watertightness at the upstream and downstream faces. A porous upstream face and lift joints in conjunction with an impermeable downstream face may result in a pressure gradient through a cross section of the dam considerably greater than that outlined above for conventional concrete. Wave Pressure The portions of the dam is subjected to the impact of waves, which is produced in water surface during wind

Figure G.D. 8 The wave pressure is tdetermined by the following formula developed by D. A. Molitor Wave height, hw = 0.032(V.F.)1/2+0.763 0.271(F)1/4 for F32 km ---------(B) where hw = height of wave in metres from top of crest to bottom of trough. F = fetch or straight length of water expanse in km V = velocity of wind in km.hr The maximum pressure intensity due to wave pressure is given by Pw = 2.4 x x hw and act at hw/2m above still water. Total force due to wave pressure Pw = (2.4 w. hw) x 5/3 . hw = 2 . w . h2 = 2 . 1000 .hw2 kg/m This force act at 3/8hw above the still water level of the reservoir.

2.4h

w

5/3hw hw

Water surface Pw

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Wind Load Wind load in stability analysis is usually ignored Earth and Silt Earth pressures against the dam may occur where backfill is deposited in the foundation excavation and where embankment fills abut and wrap around concrete monoliths. The fill material may or may not be submerged. Silt pressures are considered in the design if suspended sediment measurements indicate that such pressures are expected. Whether the lateral earth pressures will be in an active or an at-rest state is determined by the resulting structure lateral deformation. Earthquake Forces (1) General. (a) The earthquake loadings used in the design of concrete gravity dams are based on design earthquakes and site-specific motions determined from seismological evaluation. As a minimum, a seismological evaluation should be performed on all projects located in seismic zones. (b) The seismic coefficient method of analysis should be used in determining the resultant location and sliding stability of dams. In strong seismicity areas, a dynamic seismic analysis is required for the internal stress analysis. (c) Earthquake loadings should be checked for horizontal earthquake acceleration and, if included in the stress analysis, vertical acceleration. While an earthquake acceleration might take place in any direction, the analysis should be performed for the most unfavorable direction. (2) Seismic coefficient. The seismic coefficient method of analysis is commonly known as the pseudostatic analysis. Earthquake loading is treated as an inertial force applied statically to the structure. The loadings are of two types: inertia force due to the horizontal acceleration of the dam and hydrodynamic forces resulting from the reaction of the reservoir water against the dam (see Figure G.D.9). The magnitude of the inertia forces is computed by the principle of mass times the earthquake acceleration. Inertia forces are assumed to act through the center of gravity of the section or element. The seismic coefficient is a ratio of the earthquake acceleration to gravity; it is a dimensionless unit, and in no case can it be related directly to acceleration from a strong motion instrument. The coefficients used are considered to be the same for the foundation and are uniform for the total height of the dam.

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Figure G.D. 9. Seismically loaded gravity dam, nonoverflow monolith (a) Inertia of concrete for horizontal earthquake acceleration. The force required to accelerate the concrete mass of the dam is determined from the equation:

Pe = Max =( W/g)g = W Where Pex = horizontal earthquake force

M = mass of dam ax = horizontal earthquake acceleration = g W = weight of dam g = acceleration of gravity a = seismic coefficient

(b) Inertia of reservoir for horizontal earthquake acceleration. The inertia of the reservoir water induces an increased or decreased pressure on the dam concurrently with concrete inertia forces. FigureG.D.9 shows the pressures and forces due to earthquake by the seismic coefficient method. This force may be computed by means of the Westergaard formula using the parabolic approximation:

Pew = 2/3 Ce () y (hy)1/2 where Pew = additional total water load down to depth y (kips) Ce = factor depending principally on depth of water and the earthquake vibration period, t , in seconds e h = total height of reservoir (feet)

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Westergaard's approximate equation for Ce, which is sufficiently accurate for all usual conditions, in pound-second feet units is:

2

000,172.01

51

et

h

Ce

where t is the period of vibration. COMBINATION OF FORCES FOR DESIGN The design of a gravity dam is performed through an interative process involving a preliminary layout of the structure followed by a stability and stress analysis. If the structure fails to meet criteria then the layout is modified and reanalyzed. This process is repeated until an acceptable cross section is attained. Analysis of the stability and calculation of the stresses are generally conducted at the dam base and at selected planes within the structure. If weak seams or planes exist in the foundation, they should also be analyzed. Basic Loading Conditions Dams are designed for the most adverse combination of load conditions as have reasonable probability of simultaneous occurrence. The following basic loading conditions are generally used in concrete gravity dam designs (see Figure G.D 10).

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(1) Load Condition No. 1 - unusual loading condition - construction. (a) Dam structure completed. (b) No headwater or tailwater. (2) Load Condition No. 2 - usual loading condition - normal operating. (a) Pool elevation at top of closed spillway gates where spillway is gated, and at spillway crest where spillway is ungated. (b) Minimum tailwater. (c) Uplift. (d) Ice and silt pressure, if applicable. (3) Load Condition No. 3 - unusual loading condition - flood discharge. (a) Pool at standard project flood (SPF). (b) Gates at appropriate flood-control openings and tailwater at flood elevation. (c) Tailwater pressure. (d) Uplift. (e) Silt, if applicable. (f) No ice pressure. (4) Load Condition No. 4 - extreme loading condition - construction with operating basis earthquake (OBE). (a) Operating basis earthquake (OBE). (b) Horizontal earthquake acceleration in upstream direction. (c) No water in reservoir. (d) No headwater or tailwater. (5) Load Condition No. 5 - unusual loading condition - normal operating with operating basis earthquake. (a) Operating basis earthquake (OBE). (b) Horizontal earthquake acceleration in downstream direction. (c) Usual pool elevation. (d) Minimum tailwater. (e) Uplift at pre-earthquake level. (f) Silt pressure, if applicable. (g) No ice pressure. (6) Load Condition No. 6 - extreme loading condition - normal operating with maximum credible earthquake. (a) Maximum credible earthquake (MCE). (b) Horizontal earthquake acceleration in downstream direction. (c) Usual pool elevation. (d) Minimum tailwater. (e) Uplift at pre-earthquake level. (f) Silt pressure, if applicable. (g) No ice pressure. (7) Load Condition No. 7 - extreme loading condition - probable maximum flood. (a) Pool at probable maximum flood (PMF). (b) All gates open and tailwater at flood elevation. (c) Uplift. (d) Tailwater pressure. (e) Silt, if applicable. (f) No ice pressure. b. In Load Condition Nos. 5 and 6, the selected pool elevation should be the one judged likely to exist coincident with the selected design earthquake event. This means that the pool level occurs, on the average, relatively frequently during the course of the year.

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Stability Considerations a. General requirements. The basic stability requirements for a gravity dam for all conditions of loading are: (1) That it be safe against overturning at any horizontal plane within the structure, at the base, or at a plane below the base. (2) That it be safe against sliding on any horizontal or near-horizontal plane within the structure at the base or on any rock seam in the foundation. (3) That the allowable unit stresses in the concrete or in the foundation material shall not be exceeded. Characteristic locations within the dam in which a stability criteria check should be considered include planes where there are dam section changes and high concentrated loads. Large galleries and openings within the structure and upstream and downstream slope transitions are specific areas for consideration. b. Stability criteria. The stability criteria for concrete gravity dams for each load condition are listed in Table G.D -1.

Overturning Stability a. Resultant location. The overturning stability is calculated by applying all the vertical forces (SV) and lateral forces for each loading condition to the dam and, then, summing moments (SM) caused by the consequent forces about the downstream toe. The resultant location along the base is:

Resultant location =

V

M

b. Criteria. When the resultant of all forces acting above any horizontal plane through a dam intersects that plane outside the middle third, a noncompression zone will result. For usual loading conditions, it is generally required that the resultant along the plane of study remain within the middle third to maintain compressive stresses in the concrete. For unusual loading conditions, the resultant must remain within the middle half of the base. For the extreme load conditions, the resultant must remain sufficiently within the base to assure that base pressures are within prescribed limits.

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Sliding Stability a. General. The sliding stability is based on a factor of safety (FS) as a measure of determining the resistance of the structure against sliding. The horizontal components of the loads acting on a dam are resisted by frictional or shearing forces along horizontal or nearly horizontal planes in the body of the dam, on the foundation or on horizontal or nearly horizontal seams in the foundation. It follows that the total magnitude of the forces tending to induce sliding shall be less than the minimum total available resistance along the critical path of sliding. The sliding resistance is a function of the cohesion inherent in the materials and at their contact and the angle of internal friction of the material at the surface of sliding. Definition of sliding factor of safety. (1) The sliding FS is conceptually related to failure, the ratio of the shear strength (tF), and the applied shear stress (t) along the failure planes of a test specimen according to Equation 4-2:

c

F

F

CA

F

uw

P

cSF

tan1tan..

where tF = s tan f + c, according to the Mohr-Coulomb Failure Criterion; w = total weight of dam; u = total upthrust force; tan = coefficient of internal friction of material; c = cohesion of the material at the plane considered; A = area under consideration for cohesion; F =partial factor of safety in respect of friction; Fc = partial factor of safety in respect of cohesion and P = total horizontal force. Table G.D.2: Partial factors of safety against sliding (taken from Textbook of Water and Power Engineering, R.K. Sharma & T.K. Sharma)

Sl. No. Loading conditions

F Fc

For dams and the contact plane with

foundation

For foundation Thoroughly Others investigated

(i) 1, 2, 3 1.5 3.6 4.0 4.5

(ii) 4, 5 1.2 2.4 2.7 3.0

(iii) 6, 7 1.0 1.2 1.35 1.5

Safety against Crushing Safety against crushing is ensured if the compressive stresses produced are within the allowable stresses.

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Maximum compressive stress = Direct stress + Bending moment Bending moment MB

B

e

A

W

BA

eW

A

W .61

.

..6

where W = weight of dam, A = area of dam section, e = eccentricity ELEMENTARY PROFILE OF A GRAVITY DAM The elementary profile of the gravity dam to bear only the water pressure, will be triangular in section as shown below. The width of the profile will be zero at the water surface, where the pressure is zero and maximum at the base, where the pressure is maximum. Thus the shape of the elementary profile is the same as that of the hydrostatic pressure distribution. When the reservoir is empty, the only force acting is self weight (W) of the dam acting at a distance B/3 from the heel. It is the maximum possible inner-most position of the resultant so that no tension develops and provides the maximum possible stabilizing force against overturning without causing tension at toe under empty dam condition. If any triangular profile other than the right-angled is provided, its weight will act closer to the upstream face to provide a higher stabilizing force but will cause tension to develop at the toe. Vertical stresses developed when dam is empty will be:

heelatB

e

A

WP

toeatB

e

A

WP

.......6

1

......6

1

min

max

Now when reservoir is full and downstream empty, forces acting on elementary profile will be:

P

R W

e

h

h/3

P

B

2

B

2

x

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a) Dam weight W = B. h.sg. w where sg = specific gravity of dam material (for concrete, = 2.4) w = unit weight of water (9.81 x 1000 kg/m3) b). Water Pressure, P = .w.h2 acting at 1/3 from the base. c). Uplift pressure u = 1/2.c. w .B.h where c = uplift pressure intensity coefficient. The base width, B of the elementary profile is determined by the following two vriteria:

a) Stress criteria b) Stability or sliding criteria.

a) Stress criteria. When the reservoir is empty, there is no tension in the dam, the resultant is acting at the inner 1/3rd point J. When the reservoir is full, for no tension, the resultan must pass at outer 1/3rd point K. Now taking moment of all forces about point K, . . w h2.h/3 + .c. w.B.h.B/3 - .B.h. sg. w. B/3 =0 Multiplying both sides by 6/ w.h h2 + c.B2 B2 sg = 0 or B2(sg c) = h2 from whence B = h/( sg-c)1/2 ------------------------------------- (B)

h

3

P P

R

u

h

W

B

3

B

3

h

Uplift pressure

J

K

18

By considering the force triangle, using similar triangles, we have: (W u )/P = (h/3)/(B/3)

or

csgh

B

hcsgB

B

h

h

hBcBhsg w

22

2

2

1

...2

1

2

1

When the uplift force is not considered, c = 0 or B = h/ (sg)1/2 Stability or Sliding Criteria For no sliding of the dam, the horizontal forces causing sliding should be equal to the frictional forces, i.e. P = (W u) or 1/2w.h2 = (1/2.B.h.sg. w 1/2. c.B. w.h B = h/(sg c) And neglecting uplift pressure, B = h/sg ----------------------------------------------- The base width B of the elementary profile should be greater of the widths obtained in equation (B) or (C) Stresses in the elementary profile The normal stressing the dam is given by:

)(6

1 DB

e

B

uWp

when the reservoir is full, the normal stress at toe is

)(.....2/1..2/12

211

EcsghhBcsghBB

p

B

uW

B

uWp

www

The corresponding stress at the heel is:

19

)(011 FB

uWp

When the reservoir is empty, the only force acting on the elementary profile is its weight, acting through J. In this case, the maximum compressive stress at the heel = W/B(1+1) = 2W/B and the corresponding normal stress at toe is W/B(1-1) =0 The Practical Dam Profile The elementary profile of a gravity dam is a triangle with maximum water surface at its apex. This profile is only theoretical one. For meeting the practoical requirements certain changes have to be made namely: i) for communication, road has to be provided and therefore a top width; ii) for wave action, free board above the high flood level must be provided. The addition of the above will cause the resultant force to shift towards the heel. Earlier, when the reservoir was empty, the resultant was passing through the inner middle third. The above changes will shift it towards the heel, crossing the inner middle third point; this will create tension in the toe. To prevent this tension, some concrete is added in the dam body towards upstream side.

Figure G.D. : Practical dam profile

Free board

a

h Dam

a

16

20

Principal and Shear Stress in Dam

Check drawing from main notes The figure above shows the dam with possible pressure distribution on it. The maximum normal stress in the dam is the major principal stress which will be generated on the major principal plane. When the dam is full, the vertical direct stress is the maximum at the toe as the resultant is near the toe. The principal stresses near the toe is shown on the second diagram with a small element ABC. Let dr, ds and db be the lengths of AB, AC and BC respectively, and let p = be intensity of water pressure 1 = principal stress on plane AB

= shear stress Now considering unit length of the element ABC of the dam, the normal forces on the planes AB, AC and BC are 1.dr 1.dr, pv.db and p.ds sin respectively. Resolving all the forces in the vertical direction,

p v.db = p.ds sin + 1.dr.cos ----------------- (G) But dr = db.cos and ds = db.sin Therefore pv .db = p.ds.sin 2 + 1.db.cos 2 Or pv = p.sin 2 + 1.cos 2

Need to redraw

21

Therefore 1 = (p v p.sin 2 )/cos 2 1 = p v. sec 2 p.tan 2 ---------------------- (H) Equation (H) is known as the principal stress relationship, which is applicable to both upstream and downstream faces. For the downstream side the worst condition will be when there is no tail water, and hence p will be zero. In this case the major principal stress 1 is given by 1 = p v. sec 2 -----------------------------(I) If pe is the intensity of hydrodynamic pressure of tail water due to an earthquake, the principal stress at the downstream is given by: 1 = p v. sec 2 (p pe.) tan 2 In the same way,considering the hydrodynamic pressure for the upstream side, in the horizontal direction;

.db = 1.dr.sin + p.ds cos = 1..sin (dr/db) + p.(ds/db)cos = (1-p).sin .cos .

Now substituting the value of 1 from eq. (H)

= (p v sec 2 - p.sec 2 ) cos .sin or = (p v - p) tan The shear stress for the upstream side has the same value but with reversed direction For the upstream side, = - (p v - p) tan STABILITY ANALYSIS OF GRAVITY DAMS The stability analysis of a gravity dam section can be done by any one of the following methods:

a) Gravity method of two-dimensional method b) Slab analogy method c) Trial load twist method d) Lattice analogy method

Gravity Method or Two-dimensional Method. Being an approximate method, it is used for the preliminary calculations. The gravity method can be carried out by:

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i) Graphical method or ii) Analytical method.

We shall consider only the analytical method. For the analytical method, the following steps are carried out:

a) Considering unit length of the dam, all vertical loads are determined the algebraic sum of all vertical forces V is calculated

b) Considering unit length of the dam, all horizontal forces are determined and their algebraic sum H calculated.

c) The sum of the righting moments (MR) and the sum of overturning moments MO) at the toe of the dam are calculated.

The difference between the algebraic sum of the overturning and righting moments is determined i.e M = MR - MO

d) The location of the resultant force FR from the toe of the dam is also calculated by the following:

)(JV

Mxe

e) Now the eccentricity of the resultant force is determined by:

)(2

KxB

e e

f) The normal stress at the toe of the dam is determined by:

)(.6

1 LB

e

B

VN T

g) The normal stress at the heel is determined by

)(.6

1 MB

e

B

VN H

h) The principal and shear stresses at the toe and heel are determined from 1 = p v. sec 2 -----------------------------(I)

= - (p v - p) tan

i) The factor of safety against overturning is calculated by:

23

O

R

M

MSF .).(

j) The factor of safety against sliding is calculated by:

H

qBVSFfactorfrictionShear

H

VSFfactorSliding

.).(..

).(.

LOW AND HIGH GRAVITY DAMS Low Dam: is one of limiting height such that the resultant of all forces passes through the middle third and the maximum compressive stress at the toe does not exceed the permissible limit i.e 1 =wH(s-c+1) = fc

or limiting height

1.

1.

gs

fH

orcgs

fH

w

w

The limiting height, for the usual stress of dam material i. e. w = 9.81 x1000 kg/m3, = 2.4 and f = 30 kg/cm2 (or 300tonnes /m2) gives:

mx

xH 99.8

14.2100081.9

1000300

s = Specific gravity of material; c = coefficient of uplift pressure, w = specific weight of water; f = permissible compressive stress of material. High Dam: A dam the height of which exceeds the limiting height of low dams is termed as high. In a high dam, the allowable stress are often exceeded if the resultant of all the forces were to pass through the middle third; to avoid excessive stresses the resultant is maintained still near the centre of the base for which purpose the downstream slope is flattened and the upstream slope is also provided with a batter.

24

Design of Gravity Dams Before starting, one must establish whether it is low or high dam using the above relationship. .

Top width is chosen according to Creager must be about 14%of the dams height.

B

Limiting height

of low dam

Height of

high dam

Upstream

batter

Figure G.D. : Low dam and high dam

Reservoir level

H1

Figure G.D. 13. Economic section of Low Gravity Dam

cS

hB

s

1

B

cSah s 21

h

a

J S

K

Q

O

25

Free Board: is provided on the basis of height of waves and other practical considerations. In practice a free board of between (3 5)% of dam height is used

Base width B1 I is given by

cSh

B

orcS

hB

s

s

1

1

WORKED EXAMPLES The diagram below shows the cross section of a masonry dam. Determine the stability of the dam. Also determine the principal stress at the toe and heel of the dam. Take unit weight of dam material as 2250 kg/m3, density of water 1000 kg/m3 and the permissible shear stress of joint = 15 kg/cm2. Assume value of coefficient of friction = 0.75 SOLUTION Stability of dam is tested assuming no free board. A. Vertical forces i) Self weight of dam = (12 + 2.5)/2 x 15 x 1 x 2250 = 244,687.5 kg

ii) Weight of water in column DDA = (1 x 15)/2 x 1000 = 7500 kg iii) Uplift force on dam =(15 x 12)/2 x 1000 = 90,000kg

D

A B

15m

C

1 2.5 8.5

D

26

iv) Therefore V = 244687.5 + 7500 90,000 = 162.187.5 kg v) Horizontal water pressure = (w x h2)/2 = (1000 x 152)/2 = 112,500 g Calculation of moments due to various forces about toe of dam vi) Moment of self weight [(1x 15)/2](1 + 2.5 + 8.5) + (2.5x 15 x 2250)[(2.5/2)+ 8.5] + [(8.5x15)/2](2/3x8.5) = 194,062.5 + 822,656.25+ 406,406.25 = 1,423,124.9kg-m (+ve) vii) Moment due to weight of water in DDA 7500 x (1 +2.5 + 8.5) = 78,750kg-m (+ve) viii) Moment due to uplift force 82,500 x 2/3 x 12 = 660,000 kg-m (-ve) ix) Moment due to horizontal water pressure 112,500 x 1/3 x 15 = 562,500 kg-m (-ve) Therefore M =1,423,124.9 + 78750-660000-562500 = + 279,374.9 kg-m Factor of safety calculation x) Factor of safety against overturning = Resisting Moments = Overturning Moments (1423124.9 + 78750) = 1.228 < 2 unsafe. (660,000 + 562,500) xi) Factor of safety against sliding = V = 0.75 x 162,187.5 = 1.08 >1.0 safe H 112,500 xii) Shear friction factor =V + b.q = 0.75x 162,187.5 + 12 x 15 x 104 = 17.08 H 112,500 Stress calculation Let the resultant be acting at xav from the toe Xav = M = 279,374.9 = 1.72m H 162,187.5 Distance of resultant force from centre of dam, (eccentricity, e) e = B/2 xav = 12/2 1.72 = 4.28 Compressive stress at toe ft =V(1+6.e/B)=162,187.5(1+6x 4.28/12) = 509,268.7 > 50kg/cm2 unsafe

27

B 12 Tensile stress at heel, fh = V(1 6.e/B)=162,187.5(16x4.28/12)= -15407.81 =-1.54 kg/cm2 unsafe B 12 In masonry dams, there should not be any tensile stress, therefore the sectionis not safe. Calculation of principal stresses From the diagram, tan = 8.5/15 = 0.567 Sec = (1 + 152)/15 = 1.002 Sec = (8.52 + 152)/15 = 1.149 tan = 1/15 = 0.067 Principal stress at toe = pn sec2 = 509,268.7 x 1.149 = 672,336 kg/m2 Shear Stress at toe = pn tan = 509,268.7 x 0.567 = 288,755 kg.m2 Principal stress at heel h = pn sec2 p. tan 2 = -15407.81 x (1.002)2 -1000 x 15 x (0.067)2 = (15469.5 - 67.33) = 15402.16 kg/m2 Shear stress at heel

= - (pn-p) tan = - [ - 15407.81 10,000] x 1/15 = -(25407.81)/15 = 1693.8 kg/m2

Example 2: From the data given below, design a stone masonry gravity dam of practical profile.

Ground level, R.L = 1130.5m

R.L of HFL = 1155.5m

Wave height = 1.0m

Specific gravity of masonry = 2.5

Permissible compressive stress for stone masonry 125t/m2

Solution

Free board height = 1.5 x height of wave = 1.5 x1.0 1.5 m

Therefore required level of top of dam = 1155.5 + 1.5 = 1157m

Height of dam = 1157 1130.5 = 26.5m

Limiting height of dam =

mx

S

f

sw

c 71.3515.2(1000

1000125

1

Therefore the dam is a low gravity dam.

28

The design of the dam can be done with respect to the details.

Depth of water = 1155.5 1130.5 = 25m

Top width of dam = 14% height of dam = 14?100 x 26.5 = 3.71m

Assume a roadway width of 4.5 m

Therefore provide top width MN of dam = 4.5m

Base width of dam PS = B1 =

mcSs

h81.15

05.2

25

; assume 16m

Extra width JS = MN/16 = 4.5/16 = 0.281m (assume 0.3m)

Vertical distance LK = 23.145.25.422 xSsa assume 14m

Vertical distance LR = mxSsa 225.25.41.31.3

Example 3

What should be the maximum height of elementary profile of a dam, it the safe limit os stress on the masonry should not exceed 350 tonnes/m2. Assume weight of masonry 2.4 tonnes/m3. Determine the base width also. Determine H and B if uplift intensity factor is 0.67 and factor of safety is 2

Solution: The limiting height of elementary profile of a masonry dam

B1 = 16m 0.3

22m

14m

25m

1130.5

4.5

1.5 1155.5

1157.0

J S P

K

R

Q

O

N M

L

DAM

29

H =

mxg

xgx

sg

f

w

c 103)14.2(1000

1000350

1

Base width B = mcSs

H66

4.2

103

ii) F.S =2, c = 0.67: H =

mxgx

xgx

cSsSxF

f

w

c 64167.04.210000.2

1000350

1.

mcSs

HB 49

67.04.2

64

Example 4.

A concrete gravity dam has maximum water level 305.0m, bed level 225.0m, top required level of dam 309.0m, downstream face slope starts at required level 300.0m, downstream slope 2:3, tail water is nil, upstream face of dam is vertical, centre line of drainage gallery is 8m downstream of upstream face, uplift pressure is 100% at heel, 50% at line of gallery and zero at toe, specific gravity of concrete is 2.4. Considering only weight, water pressure and uplift, determine

i) Maximum vertical stresses at toe and heel of dam

ii) major principal stresses at toe of dam and

iii) Intensity of shear stress on a horizontal plane near the toe

Solution:

Height of dam H = 309.0 225.0 = 84m

Depth of water h = 305.0 225.0 = 80m

1. Top width of dam = 14% of height = 0.14 x 84 = 12m

2. Bottom width of dam = 12 + (300 225) x2/3 = 62m

Calculating weight and moments by considering unit length of dam

30

Designation Force Moment arm Moment about toe

Weight of dam,

w1=12 x84 x 1 x2.4 x w

w2=50x75/2x1x2.4 w

w = w1+w2

=2419 w

=4500 w

= 6919

50 + 12/2 = 56

50 x 2/3 = 33.33

13546 w

149985 w

M1=(+)285449 w

Uplift

U1=40 x 8 x 1/2 w

U2=40 x 8 w

U3=54 x40/2

u

=160 w

=320 w

=1080 w

=1560 w

54+2/3x8=59.33

54 + 4 =58

54x2/3=36

9493

18560

38880

M2= (-) 66933 w

Water pressure

P = wh2/2=80 x80/2 w

3200 w

80 x 1/3=26.67

M3= (-) 85344 w

V = W U = 6919 1560 = 5359 w M = M1-M2-M3 = (+) 133172 w

Position of resultant from toe xav = M/V= 133172 w /5359 w=24.85

Eccentricity, e = B/2 xav 62/2 24.85=6.15m

Normal compressive stress at toe pn=V/B(1+6e/B)= 5359 w /62(1+6x6.15/62)= 138 wN/m2

Normal compressive stress at heel pn= V/B(1-6e/B)= 5359 w /62(1-6x6.15/62)= 35 wN/m2

Uplift pressure diagram

80

300

W1

W2

50

62

8

309 305

80

225

40

8 54

U1 u 2

u 1

31

Principal stress at toe = p n sec 2 p tan 2 p = zero because tail wate is zero; tan = 2/3 sec 2 = 1 + tan 2 = 1 +(2/3)2 =13/9

Therefore = p n sec 2 = 138 w x13/9=199 w N/m2

Intensity of shear stress on a horizontal plane near toe

o =( pn p)tan = (138 w 0)x2/3 = 92 w N/m2

Example 5. A concrete gravity dam has maximum reservoir level 150.0m, base level of dam = 100.0m, tail water elevation 110.0m, base width of dam 40m, location of drainage gallery 10m from upstream face which may be assumed as vertical. Compute the hydrostatic thrust and the uplift force per metre length of dam at its base level. Assume 50 % reduction in net seepage head at the location of the drainage gallery.

Solution:

Free board = 5% of dam height = 0.05x50= 2.5 adopt 3m

Dam height h = 50 + 3 =53m

Top width a = 14%xh = 0.14 x 53 = 7.5m adopt 7.5m

Designation x w Force x w Moment arm Moment about toe

X w

50

143

W1

W2

32.5

40

10

153

09 150

305

50

80

100

225

Uplift pressure diagram

30

10

30

U3 u 2

u 1

U4

32

Weight of dam

W1=7.5x53x1x2.4

W2 = 32.5 x43/2 x1 x2.4

Weight of tail water

W3 = 8x10/2x1x1

=954

=1677

40

V1 =2671

32.5+7.5/2=36.25

32.5x2/3=21.67

8 x 1/3 = 2.67

34583

36341

107

M1 (+) 71031

Uplift pressure

U1=20 x 10/2

U2=10 x30 x1

U3=10 x30 x 1

U4=20 x 30/2 x 1

100

300

300

300

V2= 1000

30 + 10 x2/3=36.67

(30 +10/2) = 35

(30/2) = 15

(30 x 2/3) = 20

3667

10500

4500

6000

M2 =24,667

Water pressure

U/S=wh2/2=wx1.10.10/2

D/S=wh2/2=w.1.10.10/2

1250

50

H3= 1200

50/3

10/3

(-) 20,833

(+)167

M3 (-) 20,666

Position of resultant from toe xav = M/V=25698/1671=15.38m

Eccentricity, e =B/2-xac=40/2 15.38 = 4.62m

Normal compressive stress at toe, pn= V/B(1 + 6e/B)

= 1671/40(1 +6.4x62/40)=70.73 w

Normal compressive stress at heel p =V/B(1 6e/B)

= 1671/40(1-6x4.62/40)= 12.82 w

Maximum principal stress at toe = pn sec2 p tan 2 p = 10 w

tan = 40/50, sec 2 = 1.64

=70.73 x 1.64 10(40/50)2 =110 w

Intensity of shear stress on a horizontal plane near toe

0 = (pn p)tan

0 = 70.73 -10)40/50 = 49 w

33

Chapter 4

SPILLWAYS AND GATES

Spillway is a passageway to convey past the dam flood flows that cannot be contained in

the allotted storage space or which are in excess of those turned into the diversion

systems. Spillways function infrequently, at times of flood or sustained high runoff, when

other facilities are inadequate. However its ample capacity is of prime importance for the

safety of the dam and other hydraulic structures. Hydraulic aspects of spillway design

relates to design of the three spillway components: control structure, discharge channel

and terminal structure. The control structure regulates outflows from the reservoir and

may consist of a sill, weir section, orifice, tube or pipe. Design problems here relate to

determining the shape of the section and computing discharge through the section. The

flow released from the control structure is conveyed to the streambed below the dam in a

discharge channel. This can be the downstream face of the overflow section, a tunnel

excavated through an abutment, or an open channel along the ground surface. The

channel dimensions are fixed by the hydraulics of channel flow. Terminal structures are

energy dissipating devices that are provided to return the flow into the river channel

without serious scour or erosion at the toe of the dam.

Spillways are usually located at the normal flood level and the water just overflows the

crest when it gets to that level. However, some spillways are equipped with gates, which

are temporary barriers installed over the permanent crest of the spillway for storing

additional water during period of low water seasons. All small flows exceeding the

barrier top level are allowed to pass over the barrier, but during large flood flows, the

barrier is removed and full spillway capacity is used to discharge the flood water.

Types of Spillways

Spillways may be classified, depending upon the type of structure as:

i) Side channel spillway ii) Straight drop spillway iii) Overflow or Ogee spillway iv) Chute or Trough spillway v) Shaft (Morning glory) spillway vi) Siphon spillway

Side channel spillway

This type of spillway is most suited for earthfill and rockfill dams, in narrow canyons,

where construction of other types of spillway is not possible. Also when a long overflow

crest is required for limiting the surcharge head, this type is suited. It is also useful when

its discharge is to be connected to a narrow discharge channel or tunnel. In this type the

control weir is kept along the side and approximately parallel to the upper portion of the

spillway. The discharge after passing over the crest turns at about 90o before flowing into

a trough to be discharged.

Straight Drop (free overfall) Spillway. In this type, water is allowed to fall freely from a

low weir and vertical fall structure. In some cases the crest of the spillway is extended in

34

the shape of overhanging lip which keeps the discharge away from the straight drop

section. Ogee (Overflow) Spillway: is a special form of a weir whose shape is made to

conform to the profile of lower nappe of a ventilated sheet of water falling from a sharp-

crested weir. The profile is so shaped that the discharging water always remains in touch

with the spillway surface.

Earthfill dam

Side channel spillway

Upstream

downstream

35

Accordingly, the profile of the ogee spillway is made to the shape of the lower nappe of a

free falling jet.

The downstream curve of the ogee has the equation:

85.085.1 2 hyx --------------------------- (S-1) where x and y are the co-ordinates of the crest profile measured from the apex of the

crest, and h is the design water head.

The upstream curve may be approximate to the following

Straight drop spillways

With downstream protection

spillway

Spillway

Without downstream protection

O

x

r1

Origin and apex of crest

y

b

r2

Design head h

a = 0.175h

b = 0.292h

r1 = 0.50h

r2 = 0.20h

a

Fig. S-1 Definition sketch of overflow spillway

36

85.0

85.127.0724.0

h

hxy

------------------------ (S-2)

For larger discharges, (eg flows beyond H.F.L), the nappe may leave the ogee profile and

will cause negative pressure resulting in cavitation and increase of discharge.

Discharge over an Overflow Spillway

The discharge equation of the ogee-shaped spillway is given by

Q = C.Le.(he)3/2 ------------------------------(S-3)

Where, Q = discharge over the ogee spillway

Le = effective length of the crest

he = h + v2/2g = total water head at crest including the velocity approach

C = a variable coefficient of discharge, whose value varies from 2.1 to 2.5

depending on various factors.

The effect length of the crest is given by the equation:

Le = L 2(NKp Ka) ---------------------------(S-4) Where L = Total clear length of crest

N = No. of piers in the spillway

Kp = pier contraction coefficient

Ka = abutment contraction coefficient

Table S-2 Pier Contraction Coefficient Kp

Table S-3 Abutment Contraction Coefficient (Kp)

Condition of Abutment Value of Ka

1 Rounded abutment with head wall as 90o to the direction

of flow, when 0.5h > r > 0.15h

2 Rounded abutment where r> 0.5Ho and the angle of head

wall with direction of flow is

37

1. Assume C = 3.95

2. from the discharge equation 1.17

32.7018095.3

000,502/3

e

e

h

CL

Qh

3. Depth of water upstream = 800 680 = 120 ft. Velocity of approach vo = 50,000/120(180) = 2.31ft/sec

Velocity head = v2/2g = (2.31)2/(32.2) = 0.08 ft

4. Maximum water head = 17.1 0.08 = 17.0ft 5. Height of crest, P = 120 17.0 = 103ft. 6. Since he 1.33, high overflow section 8. Downstream quadrant of the crest shape

y/12 = (x/12)1.85 or y = 0.06x1.85

X (select)ft Y (computed) ft

5 1.18

10 4.25

15 9.00

20 15.30

30 32.40

9. Point of tangency assume a downstream slope of 2:1 (XDT)/h = 0.485(K)1.176 where XDT = horizontal distance from the apex to the downstream tangent point

slope of the downstream face XDT = 0.485[2(2)]

1.176 (12) = 30 ft

10. Upstream quadrant. A/h = 0.28, B/h = 0.165 A = 0.28(12) = 3.36; B = 0.165(12) = 2.00ft

(x2)/(3.36)2 + (2.0 y)2/(2.0)2=1

X (selected) ft Y (computed) ft

1.0 0.09

2.0 0.39

3.0 1.10

3,36 2.00

CHUTE OR TROUGH SPILLWAY

Chute spillway is a type of spillway in which the discharge is conveyed fro a reservoir to

the downstream river level through a steep open channel placed either along the dam

abutment or through a saddle. In a chute spillway, the velocity of flow ia always greater

than the critical. The name chute applies, regardless of the control device used at the

38

head of the chute, which can be an overflow crest, a gated orifice or a side channel crest.

The chute spillway consists of four parts: an entrance channel, a control structure or crest,

the sloping chute and a terminal structure. The entrance structure is usually an open

channel of sub-critical flow. The critical velocity occurs when the water passes over the

control.

Flow in the chute is maintained at supercritical stage until the terminal structure DE is

reached. It is desirable that from B to C, where a heavy cut is involved, the chute is

placed on a light slope. From C to D it follows the steep slope and ends with an energy

dissipating device placed at the bottom of the valley D. The axis of the chute is kept as

straight as possible; otherwise, the floor has to be super-elevated to avoid the piling up

high-velocity flow around the curvature.

It is preferable that the width of the control section, the chute, and the stilling basin are

the same.

To prevent hydrostatic uplift under the chute, a cutoff wall is provided under the control

structure and a drainage system of filters and pipes is provided. When the stilling basin is

operating, there is substantial uplift under the lower part of the chute and upstream part of

the stilling basin floor. The floor must be made sufficiently heavy or anchored to the

foundation.

Slope of Chute Channel

It is important that the slope of the chute in the upstream section BC chute, should be

sufficiently steep to maintain a supercritical flow to avoid the formation of a hydraulic

jump. Therefore slope should always be above the critical slope.

Fro the Mannings formula;

Chute section

Terminal Structure

Entrance cha. Control structure

Fig. S 3 Chute Spillway Section

C B

A

D E

F

39

2.13/21

IARn

Q -------------------------- ( S-5 )

For a rectangular channel under critical flow conditions,

h3 = q2/g ----------------------------------(S-6 )

By putting this into Mannings formula, the critical slope could be determined. The slope to be given to the chute must be greater than this critical slope.

A review of existing spillways indicate that the actual slopes of the upstream section of

the chute are 1 or 2 % or more.

Chute Sidewalls

Except for the diverging or converging sections, chute channels are designed with

parallel vertical sidewalls, commonly of reinforced concrete 30 to 45cm thick; designed

as retaining wall. The height of the walls is designed to contain the depth of flow for the

spillway design flood by gradually flows hydraulics equation. Free board or allowance is

made for pier end waves, roll waves and air entrainment. In view of uncertainties

involved in the evaluation of surface roughness, pier end waves, roll waves and air

entrainment, the following empirical equation is added to the computed depth of water

surface profile

Freeboard (ft) = 2.0 + 0.002V h1/3 --------------------- (S-7)

where V - mean velocity in chute section under consideration, h mean depth.

SIDE CHANNEL SPILLWAYS

In side-channel spillway, the overflow weir is placed along the side of the discharging

channel, so that the flow over the crest falls into a narrow channel section (trough)

opposite the weir, turns through a right-angle, and continues I the direction approximately

parallel to the weir crest

40

This type of weir is adoptable to certain special conditions, such as when a long overflow

crest is desire but the valley is narrow, or where the overflow are most economically

passed through a deep narrow channel or a tunnel. The crest is similar to an overflow or

ordinary weir section. A control section downstream from the trough is achieved by

constricting the channel or elevating the channel bottom to induce critical flow.

Downstream from the control section functions as a chute spillway. Thus the side-

channel design is concerned only with the hydraulic action in the trough upstream of the

control section, where varied flow takes place. The water surface profile in the trough is

determined from the momentum equation applied to gradual varied flow. If flow rate and

velocity in left is Q1 and V1 while those at the right is Q2 and V2, then by applying the

momentum equation, the rate of change of momentum in the reach dx is equal to the

external forces acting in the reach.

Rate of momentum at left(upstream) = Q1V1 Rate of momentum at downstream = Q2V2 Change in momentum = (Q2V2 Q1V1) ------------------------------------------(S -8) Due to smallness of friction forces and weight component in the direction of flow, the

only force acting is the hydrostatic pressure force.

The resultant hydrostatic force Fp1 Fp2 = .(A1. h1 A2h2) = (Q1/V1 Q2V2)dy ---(S-9)

Trough

Side-channel spillway: plan and section

A

Dam

Side channel crest

Control section

Side channel trough

Chute

A

41

Equating equations (S-8) and (S-9) and rearranging

2

12121

2

2

1

12

Q

QQVVV

Q

V

Q

V

g

Qdy ---------------------------(S-10)

Equation (S-10) is solved by a trial and error procedure .

Worked example:

Design a side-channel trough for a spillway of 100ft length for a maximum discharge of

2500 cfs. The side-channel trough has a length of 100ft and a bottom slope of 1ft in 100

ft. A control section of 10 ft width is placed downstream from the trough with the bottom

of the control at the same elevation as the bottom of the trough floor at the downstream

end.

Solution

1/. Critical depth at the control section, yc = (q12/g)

2. q1 = 2500/10 = 250cfs/ft.

3. yc = (2502/32.2)1/3 = 12.44ft

4. vc = q1/yc = 250/12.44 = 20.1ft/sec

5. Velocity head, hc = vc2/2g = (20.1)2/2(32.2) = 6.27ft

6. For a side-channeltrough, assume a trapezoidal section with 2V:1H slope and a 10ft

bottom width. Also, assume that the transition loss from the end of side-channel trough to

the control section is equal to 0.2 of the difference in velocity heads between the end of

the transition.

dx

P2 P1

dy

Crest

42

7. The following energy equation may be written between the trough end and the control

section;

y100 + h100 = yc + hc + 0.2(hc h100) or y100 + 1.2h100 = 12.44 + 1.2(6.27) = 19.96 --------- (S-11)

where the subscript c refers to critical and 100 refers to the distance of the trough from the upstream end of the spillway. Equation (S-11) is solved by trial and error.

Assume y100 = 19.2; then A = 376.3 ft2, v = Q/A = 2 500/376.3 = 6.64 ft/sec, and h100 =

6.64)2/2(32.2)0.68. Thus Eq. (S-11) is satisfied.

8. The known values above relate to section 1 at the downstream end of the trough.

Section 2 is taken at the upper end of a selected increment dx, 25 ft in this case. A value

of the change in water level, dy, is assumed for each reach, all terms of equation (S-10)

are evaluated, and dy is computed, as in table below. The assumed and computed values

of dy should match; otherwise, a new value is assumed for dy.

9. The process is repeated until the upstream end of the channel is reached.

1 2 3 4 5 6 7 8

dx

(selected)

Bottom

Level

dy

(Assume)

Water

Level

y

(col 4-col 2)

A Q =

q(L-dx)

v

= Q/A

D/S end 100.0 119.2 19.20 376.3 2500 6.64

25 100.25 1.0 120.2 19.95 398.5 1875 4.71

0.62 119.82 19.57 387.2 4.84

25 100.50 0.50 120.32 19.82 394.6 1250 3.17

0.40 120.22 19.72 391.6 3.19

25 100.75 0.25 120.47 19.72 391.6 625 1.60

15 100.90 0.10 120.57 19.67 390.2 250 0.64

0.07 120.54 19.64 389.3 0.64

9 10 11 12 13 14

Q1 + Q2

Q2 Q1 v1 + v2 v2 v1 dy computed

Remarks on

assumed dy

4375 -625 11.35 -1.93 0.63 High

4325 -625 11.48 -1.80 0.61 Ok

3125 -625 8.01 -1.67 0.41 High

3125 -625 8.03 -1.65 0.41 Ok

1875 -625 4.79 -1.59 0.24 Ok

875 -375 2.24 -0.96 0.07 High

875 -375 2.24 -0.96 0.07 ok

Note: 1. q = Q/L = 2500/100= 25cfs/ft ; bottom slope = 1 in 100ft given

2. Bottom level (col. 2) = slope x channel length + datum

3. Water level (col. 4) = final water level at the preceding station (section) + assumed dy.

4. A (col. 6 = area of cross-section of trough computed for depth y in col.5

5. Q (col.7) = q(L - dx), L = crest length 6. Q1 + Q 2 (col. 9) = Col. 7 + value in col. 7 at the preceding station (section)

43

MORNING GLORY OR SHAFT SPILLWAYS

In this type of spillway, water enters over a horizontal circular crest and then drops

through a circular shaft, and then through a horizontal tunnel or conduit. This type of

spillway is suitable under the following situations:

1. For damsites with steeply rising abutments particularly where a diversion tunnel can be utilized as discharge carrier.

2. For sites in narrow canyons. 3. For sites where there is inadequate space for locating other type of spillway.

Advantages: The main advantage of such type of spillways are as follows:

1. The nearly maximum capacity may be attained at relatively low heads. 2. It is ideal for sites where maximum spillway overflow is to be limited.

This type of spillway consists of four parts: 1) a circular weir at the entry, 2) a flared

transition conforming to the shaft of the lower nappe of a sharp-crested weir, 3) a vertical

drop shaft, and 4) a horizontal or near-horizontal outlet conduit or tunnel. As the head

increases, the control shifts from weir crest, to drop shaft, and to outlet conduit.

Table S-3. Discharge characteristics of shaft spillway

Control Point Condition Characteristics Discharge relations

Weir crest Unsubmerged flow Weir flow Q = CLh3/2 --------(S-12)

Throat of drop

shaft

Partially

submerged

Orifice flow 95.0...21 dad CgHAcQ

----------------- (S-13)

Downstream of

outlet conduit

Submerged flow Pipe flow

K

gHAQ T

22 -----(S-14)

K = loss coefficient through pipe

Condition 1: a free-discharging weir prevail as long as the nappe forms to converge into

the shape of a solid jet.

Condition 2: weir crest is drowned out. The US Bureau of Reclamation (1977) indicated

that this condition is approached when Hd/Rs >1, where Hd is the design head and Rs is

the radius of the crest.

Condition 3: Spillway is flooded out, showing only a slight depression and eddy at the

surface. Under condition 3, the head rises rapidly for a small increase in discharge. Thus

the design is not recommended under this condition (i.e., under the design head, the outlet

conduit should not flow more than 75% full).

The U.S. Bureau of Reclamation suggested that the following weir formula may be used

for flow through the shaft spillway entrance regardless of the submergence, by adjusting

the coefficient to reflect the flow conditions.

Q = C(2Rs)H3/2 ----------------------------- (S-15)

44

Where C = discharge coefficient related to Hd/Rs and P/Rs , where Hd = design head and

P = crest height from the outlet pipe; Rs = radius to the circular crest; H = head over the

weir ( Gupta page 495)

Equation (S-15) can be used to determine the crest size (radius), Rs for a given discharge

under the maximum head.

Equation (S-13) can be used to determine the shape of the transition (drop shaft) that is

required to pass the design discharge with the maximum head over the crest.

Worked Example: A shaft spillway is to discharge 2000 cfs under a design head of 10 ft.

Determine the minimum size of the overflow crest. Also determine the shape of the

transition if the control section is 4 ft below the crest level.

Solution

1. Since the coefficient C is related to P and Rs, assume that P/Rs > 2 and determine Rs by trial and error.

2. Try Rs = 7 ft. Hd/Rs = 10/7 = 1.43

From the curve of C as a function of Hd/Rs, we choose C = 1.44

3. From equation (S-15)

Q = C(2Rs)H3/2 = 1.44(2) (7) (10)3/2 = 2002cfs This ois practically the same as the required discharged. Hence the crest radius = 7 ft.

4. Depth of the control section from the water surface; Ha = 10 + 4 = 14 ft.

5. From equation (S-13), 95.0...21 dad CgHAcQ

R2 = 2000/{0.95(2gHa)1/2}

or R = 9.14/(Ha)1/4

SIPHON SPILLWAYS

A siphon spillway is a short enclosed duct whose longitudinal section is curved. When

flowing full, the highest point in the spillway lies above the liquid level in the upstream

reservoir, and the pressure at that point must therefore be sub-atmospheric. This is the

essential characteristics of a siphon. Siphon spillways can be saddle type or volute type

but the volute is not very common. The siphon has usually three parts, 1) the inlet or

mouth; 2) throat, and 3) lower limb.

Ha (select), ft R (ft)

14 4.73

16 4.57

18 4.44

20 4.32

45

The siphon spillway functions as follows:

When the water level exceeds the crest level, the water commences to spill and flows

over the downstream slope in much the same way as a simple Ogee spillway. As the

water rises further, the entrance is sealed off fro the atmosphere. Air is initially trapped

within the spillway, but the velocity of flow of water tends to entrain the air (giving rise

to aeration of the water) and draws it out through the exit. When all the air has been

expelled, the siphon is primed and is therefore acting as a simple pipe. \There are thus

three possible operating conditions depending on upstream depth.

1. Gravity spillway flow 2. Aerated flow (during self priming) 3. Pipe flow (after priming).

Operational problems with siphon spillways. The aerated condition is unstable and is

maintained while the siphon begins to prime. In a simple siphon, a small change in H

produces a sharp increase or decrease in the discharge through the spillway. Depending

on the discharge entering the reservoir, the siphon could go through the following cycle:

a) if the spillway is initially operating with gravity flow, then the upstream (reservoir) level must rise,

b) when the upstream level has risen sufficiently, the siphon primes and the spillway discharge increases substantially,

c) the upstream level falls until the siphon de-primes and its discharge drops. The cycle (a) to (c) is then repeated. This obviously can give rise to radical surges and

stoppages in the downstream flow.

Other potential problems encountered with siphon spillway are:

i) blockage of spillway entrance by debris ( this problem could be overcome by submerging the inlet of the hood into the water or installing a trash-

intercepting grid in front of the intake).

ii) Substantial foundations required to resist vibrations during operation of siphon.

iii) Waves arriving in the reservoir during storms may alternately cover and uncover the entry, thus interrupting smooth siphon action

The discharge of a saddle siphon can be calculated from the following formula;

gHCAQ 2 --------------------------- (S-16)

where A = area of cross section at crown; A = L x b

L = length of hood (going into the paper)

b = height of throat

H = operating head

H = Reservoir level downstream tail water level if outlet is submerged H = Reservoir level - downstream centre of outlet if the outlet is discharging

freely.

C = Coefficient of discharge. Its average value may be taken as 0.65.

46

SPILLWAY OR CREST GATES

Spillway gates are the temporary barrier installed over the permanent crest of the

spillway, for storing additional water during dry weather season. The small flows in

excess above the spillway gates is allowed to pass over the gates, but in case of large

flood, the spillway gates are opened and the full capacity is used to remove excessive

flood water.

Spillway gates can be provided on all types of spillways except siphon spillway.

Following types of spillway gates are commonly used:

a) Flash board gates

b) Stop logs or needle gates

c) Radial gates

d) Drum gates

e) vertical lift gates

H2

Crown

Throat

Crest

De-primer

Air vent

Hood

H1

Mouth

Figure S- . Siphon Spillway

47

Flash Board Gate

Flash Board

Rubber stopper

Spillway crest

Stop logs or needle gates

Piers

Groove

Stop log